Question

If the sum of the roots of the equation $\lambda x^{2}+2 x+3 \lambda=0$ be equal to their product, then $\lambda$ =
A. 4
B. -4
C. 6
D. $-\dfrac{2}{3}$

Answer 1

Hint: We need to find the value of $\lambda$. We are given that the sum and products of the roots of the equation $\lambda x^{2}+2 x+3 \lambda=0$ are equal. To solve this question recall the relationship between the roots and the coefficients of the equation.

Formula Used: Let $a x^{2}+b x+c$ be a polynomial having $\alpha$ and $\beta$ as the zeros of the polynomial. Then,
Sum of the roots of the polynomial, $\alpha+\beta=-\dfrac{\text { coefficient of } x}{\text { coefficient of } x^{2}}= -\dfrac{b}{a}$
Product of the roots of the polynomial, $\alpha \beta=\dfrac{\text { constant term }}{\text { coefficient of } x^{2}}=\dfrac{c}{a}$

Complete step by step solution: We have the equation $\lambda x^{2}+2 x+3 \lambda=0$. Let $\alpha$and $\beta$ be the roots of this equation. We are given that the sum and products of the roots are equal.
That is, $\alpha+\beta=\alpha \beta$
We have, $\alpha+\beta= -\dfrac{\text { coefficient of } x}{\text { coefficient of } x^{2}}= -\dfrac{b}{a}$ and $\alpha \beta=\dfrac{\text { constant term }}{\text { coefficient of } x^{2}}=\dfrac{c}{a}$
Therefore, $\alpha+\beta=-\dfrac{2}{\lambda}$ and $\alpha \beta=\dfrac{3 \lambda}{\lambda}=3$
Now, $-\dfrac{2}{\lambda}=3$
$\implies \lambda=-\frac{2}{3}$

So, Option ‘D’ is correct

Note: Almost all questions that are related to roots and coefficients of the quadratic equation can be solved by their relationships. But be careful with the operations we perform, for example, in this question is in the denominator of the fraction you may mess up at this step.